python – Use a.any() or a.all()

python – Use a.any() or a.all()

If you take a look at the result of valeur <= 0.6, you can see what’s causing this ambiguity:

>>> valeur <= 0.6
array([ True, False, False, False], dtype=bool)

So the result is another array that has in this case 4 boolean values. Now what should the result be? Should the condition be true when one value is true? Should the condition be true only when all values are true?

That’s exactly what numpy.any and numpy.all do. The former requires at least one true value, the latter requires that all values are true:

>>> np.any(valeur <= 0.6)
>>> np.all(valeur <= 0.6)

There is one more way you can get this

import numpy as np

a = np.array([1,2,3,4])
b = np.array([5,6,7,8])
c = np.array([1,2,3,4])

print((a == b ).all())  #False
print((a == c ).all())   # True
print((a == b ).any())   #False
print((a == c ).any())   #True
print((a > 3 ).all())    #False

python – Use a.any() or a.all()

You comment:

valeur is a vector equal to [ 0. 1. 2. 3.] I am interested in each single term. For the part below 0.6, then return this works….

If you are interested in each term, then write the code so it deals with each. For example.

for b in valeur<=0.6:
    if b:
        print (this works)
        print (valeur is too high)

This will write 2 lines.

The error is produced by numpy code when you try to use it a context that expects a single, scalar, value. if b:... can only do one thing. It does not, by itself, iterate through the elements of b doing a different thing for each.

You could also cast that iteration as list comprehension, e.g.

[yes if b else no for b in np.array([True, False, True])]

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